Why is eπ√163e^{\pi \sqrt{163}} almost an integer?

This is quite a challenge to express in “layman’s terms”, but the
reason is that
$$j\left(\frac{1+\sqrt{-163}}{2}\right)$$
is an integer where $j$ is the $j$-function. When you substitute
$(1+\sqrt{-163})/2$ into the $q$-expansion (see the wikipedia page)
of $j$, all terms save the first two are small, and the first two equal
$$-\exp(\pi\sqrt{163})+744.$$

The reason that this $j$-value is an integer is due to the quadratic
field $\mathbb{Q}(\sqrt{-163})$ having
class number one, or equivalently that all positive-definite
integer binary quadratic forms of discriminant $-163$ are equivalent.

Added
I’ll try to explain the connection with binary quadratic forms. Consider
a quadratic form
$$Q(x,y)=ax^2+bxy+cy^2$$
with $a$, $b$ and $c$ integers. I’ll only consider forms $Q$ which are
primitive, so that $a$, $b$ and $c$ have no common factor $> 1$,
and positive-definite, that is $a > 0$ and the discriminant
$D=b^2-4ac < 0$. There is a notion of equivalence of quadratic forms,
and two primitive positive-definite forms $Q$ and $Q'(x,y)=a'x^2+b'xy+c'y^2$
(necessarily also of discriminant $D$) are equivalent if and only if
$$j\left(\frac{b+\sqrt{-D}}{2a}\right) =j\left(\frac{b'+\sqrt{-D}}{2a'}\right).$$
For each possible discriminant there are only finitely many equivalence
classes. Thus we get a finite set of $j$-values for each discriminant, and
the big theorem is that they are the solutions of a monic algebraic equation
with integer coefficients. When there is only one class the equation has the
form $x-k=0$ where $x$ is an integer, and the $j$-value must be an integer.

My recommended reference for this is David Cox's book
Primes of the form $x^2+ny^2$. But these results appear towards
the end of this 350-page book.